In your code, circuits is an array of array of quantum circuits and each element of circuits is an array of quantum circuits. In the code, you are using a for loop to run each element of the circuits ...

Say you have two bases $B_1$ and $B_2$. Let $U$ be the unitary that transforms the orthonormal basis vectors in $B_1$ to the basis vectors in $B_2$. So, it is obvious that $U^{\dagger}$ will be the ...

If you recall from the analysis of Grover's algorithm, if the success probability of obtaining an item of interest is $\sin^2(\theta)$ just before a Grover iteration, then just after one iteration of ...

Contrary to what you have mentioned in the question, when you use a $u_3$ gate, Qiskit does not break it up into multiple gates. In fact $u_3$ gate is one of the basis gates for all Qiskit backends ...

Short Answer: There is no universal gate that can entangle the output state for any given ? and ??? operators in the circuit given in the question. Long Answer: Say you have the entangle state $|\psi\... View answer Accepted answer 3 votes We know that operators are applied on qubits. But sometimes it does not make sense to ask the state on which we apply an operator, say$U$. For instance, take the two-qubit entangled state$|\psi\...

Measuring in the 'Computational Basis' always means to measure in the $\{|0\rangle,|1\rangle\}^n$ basis for $n$ qubits. As for your second question, I feel that you misinterpreted the author. It is ...

I assume that you have a qubit register $q$ and given that the state of $q$ is $|\psi_i\rangle$ you want to apply $U_i$ to $q$ for $i=0,1,2$. If this is what you wish to do, then unfortunately if the ...

To add to the circuit by @Craig, the reason why papers tend to mention that some gate is obtainable by $\sqrt{swap}$ upto a few single qubit gates is because the set that contains the $\sqrt{swap}$ ...

If I am correct, I suppose you are talking about the Qiskit Challenge 2020. A possible reason why your circuit is being graded wrong is because the question asks you to construct the circuit for full ...

It is correct. Since you have $R_1(\theta) = \begin{bmatrix}1 & 0\\ 0 & e^{i\theta}\end{bmatrix}$ and $R_z(\theta) = \begin{bmatrix}e^{-i\theta/2} & 0\\ 0 & e^{i\theta/2}\end{bmatrix}$ ...
Quantum phase estimation does not have anything to do with $\theta$. I feel that you are confusing phase estimation with the implementation of HHL as given in the paper https://arxiv.org/abs/1804....